The fastest recorded pitch was clocked at 160 km/h. If a pitch were thrown
horizontally with this velocity, how far would the ball fall vertically by the time it
reached home plate, 20 m away?
To solve this problem, we need to calculate the time it takes for the ball to travel 20 m horizontally at a velocity of 160 km/h.
First, we convert the velocity from km/h to m/s:
160 km/h = 160,000 m/h
160,000 m/h / 3600 s/h = 44.44 m/s
Next, we use the equation for horizontal distance:
horizontal distance = velocity * time
20 m = 44.44 m/s * time
time = 20 m / 44.44 m/s
time = 0.45 s
Now that we have the time it takes for the ball to reach home plate horizontally, we can calculate the vertical distance it falls using the equation for vertical distance:
vertical distance = 0.5 * gravity * time^2
vertical distance = 0.5 * 9.8 m/s^2 * (0.45 s)^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2
vertical distance = 0.5 * 9.8 m/s^2 * 0.2025 s^2 = 0.44775 m
Therefore, the ball would fall vertically by approximately 0.45 m by the time it reaches home plate.